theorem
  g + h = h + g implies g + ( i * h) = i * h + g
proof
  assume
A1: g + h = h + g;
  thus g + ( i * h) = 1 * g + ( i * h) by Th25
    .= i * h + ( 1 * g) by A1,Th38
    .= i * h + g by Th25;
end;
